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## Differential Calculus

### Course: Differential Calculus > Unit 1

Lesson 16: Intermediate value theorem- Intermediate value theorem
- Worked example: using the intermediate value theorem
- Using the intermediate value theorem
- Justification with the intermediate value theorem: table
- Justification with the intermediate value theorem: equation
- Justification with the intermediate value theorem
- Intermediate value theorem review

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# Justification with the intermediate value theorem: equation

Example justifying use of intermediate value theorem (where function is defined with an equation).

## Want to join the conversation?

- 3/4 is 0.75 and 1/2 is 0.5 right ? so how is 3/4 between 1 and 1/2 ? am i just being dumb ?(10 votes)
- think about it like this:

f(1) = 1/1 = 1

f(2) = 1/2 = 0.5

3/4 = 0.75 must fall between 0.5 and 1!

In fact, it is 1/(4/3), and 4/3 is clearly between the interval [1,2], since 4/3 is ~1.33.(14 votes)

- Does the IVT work for continuous functions only?(4 votes)
- The IVT only applies to intervals where the function is continuous, otherwise, the function might skip some intermediate value and the theorem would be false.(5 votes)

- At3:03, Sal said that g(x)=3/4 is between g(1)=1 and g(2)=2.I want to know how g(x)=3/4 is between them by calculation,is it like 4/3=1.33 when it shuffle from 3/4 to 4/3?(4 votes)
- You probably have made a small mistake. At3:03, Sal said that
**g(2)=1/2**,*not g(2)=2*.

g(x)=3/4 is between 1/2 and 1. Fair enough.(2 votes)

- At2:14is there a specific reason why he included the x not being equal to 0? Perhaps my lack of understanding what defined means is part of my confusion.(2 votes)
- The interval in question is from -1 to 1, which includes 0, and at 0 the function is undefined, so that's how you establish that it isn't continuous.(4 votes)

- I think something misses to the answer at0:36. What would you do if you do happen to be able to demonstrate it through the intermediate value by selecting a shorter range in which there are not discontinuities. Example: with that same equation, in the case of being asked a solution for f(c) = 3/4 between -1 <= c <= 2, you could just "cut out" the portion of the function that interests you, one without discontinuities, such as 1 <= c <= 2 and demonstrate it.(3 votes)
- What do you mean 1/x is not continuous over that interval? You did not even substitute anything into the equation..(2 votes)
- You can determine continuity in this context by simple inspection.

x=0 => undefined.(2 votes)

- This notation was really confusing, at least for me. When I see "the equation g(x) = 3/4", I just think it denotes a straight line where y = 3/4. I didn't realize it was a potential answer to the original g(x) at the top until I read the comments. Is there a less ambiguous way of expressing this, or do I just need to get better at reading these things?(1 vote)
- It's all in the wording. If g(x) = 3/4 was given without context then yes, I'd think of a line passing through y = 3/4 too. But here, the question asks for a solution to g(x) = 3/4 given g(x) = 1/x. So, it means that you're given a function's value (3/4) at a point (which is unknown) and you're asked to find if IVT can be used to determine whether there is some x for which 1/x = 3/4 with the x being between 1 and 2, and Sal goes onto prove the same.(1 vote)

- 1:08

Wanted to fix that the function 1/x is not discontinuous at the range [-1,1], but rather only discontinuous at 0. Wrong part: [-1,1] implies discontinuity at 0 but also implies discontinuity at [-1,0)U(0,1] which is wrong.(1 vote)

## Video transcript

- [Instructor] Let g
of x equal one over x. Can we use the intermediate value theorem to say that there is a value c, such that g of c is equal to zero, and negative one is
less than or equal to c, is less than or equal to one? If so, write a justification. So in order to even use the
intermediate value theorem, you have to be continuous
over the interval that you care about. And this interval that we
care about is from x equals negative one to one. And, one over x is not
continuous over that interval, it is not defined when x is equal to zero. And so, we could say, no, because, g of x not defined, or I
could say not continuous. It's also not defined on
every point of the interval, but let's say not continuous
over the closed interval from negative one to one. And we could even put in
parentheses not defined, at x is equal to zero. All right, now let's start
asking the second question. Can we use the intermediate
value theorem to say that the equation g of x is equal to 3/4 has a solution where 1 is
less than or equal to x, is less than or equal to two? If so, write a justification. All right so first let's
look at the interval. If we're thinking about the
interval from one to two, well, yeah, our function
is going to be continuous over that interval, so we
could say g of x is continuous on the closed interval from one to two. And if you wanted to put
more justification there, you could say g defined
for all real numbers, such that x does not equal zero. I could write g of x
defined for all real numbers such that x does not equal to zero, and you could say rational
functions like one over x, are continuous at all
points in their domains. That's going really
establishing that g of x is continuous on that interval. And then we wanna see what
values does g take over, take on at the end point, or actually, these are the end points we're
looking at right over here. G of one is going to be
equal to one over one is one, and g of two is going to be one over two. So, 3/4 is between g of one and g of two, so by the intermediate value theorem, there must be an x that is in the interval from where it's talking about
the interval from one to two, such that g of x is equal to 3/4. And so, yes, we can use the
intermediate value theorem to say that the equation
g of x is equal to 3/4 has a solution, and we are done.